# 10 x 10 grid

Investigate further.

The purpose regarding this investigation is in order to prove or disprove of which there is a correlation between the products of nook numbers of any size container within any size grid. I shall calculate the diagonal difference (d) regarding a box. There are really two ways to estimate the difference between typically the products:

Versus

W

Con

Z

i actually. W x Y – V x Z

ii. V by Z – W x Y

I actually shall begin with the 10 x 10 grid, because shown above, and a new 2 x 2 box.

2 times 2 box #1

I have replaced numbers for the characters inside my two formulae.

1

a couple of

11

12

i. d = 2 x 11 – 1 x 12 sama dengan 22 – 12 = ten

2. d =1 x 13 – 2 x eleven = 12 – 22 = -10

d = +/-10

I will employ formula i, because method ii creates negative numbers, which could make the calculations more complex as compared to necessary.

2 x 2 box #2

12

13

22

twenty-three

d = 13 x 22 – 12 x 23

d = 286 – 276

d = 10

I think that will d = 10 for any 2 x a couple of box within a 10 x 10 grid. I will test this theory once again, using the highest figures possible in the main grid, to evaluate that it is usually not a coincidence.

2 x 2 box #3

89

90

99

100

d = 90 times 99 – 89 x 100

m = 8910 – 8900

d = 10

Algebraic proof: formula #1:

+1

+10

a new

a & 1

a new + 10

a + 10 & 1

m = (a + 1) (a + 10) – a (a + ten + 1)

= a2 + 10a + a + 10 – a2 – 10a – a

= 10

I have noticed that d = 10, and the grid is 12 columns (G) wide. To be able to test if this is a coincidence, I have got amended my formula.

Algebraic proof: formula #2:

+1

+G

a

a + just one

a + G

an and up. G + one

d = (a + 1) (an and up. G) – a (a + G + 1)

= a2 + aG+ a + c – a2 – aG – the

= G, exactly where G = 10, m = 10

I will test method #2 using a two x 2 box upon a different size grid. I will make use of a five x 5 grid.

2 x a couple of box in a few x 5 grid

1

2

three or more

4

5

6

7

8

9

10

eleven

12

13

14

fifteen

16

17

18

19

20

21

22

twenty-three

24

25

5 x 5 main grid

G = 5

two x 2 box #4

1

2

6

several

d sama dengan 2 x 6 – 1 x 7

d = twelve – 7

d = 5

2 x 2 box #5

7

eight

12

13

d = 8 by 12 – 7 x 13

m = 96 – 91

d = 5

two x 2 box #6

19

20

24

twenty five

d = 20 x 24 – 19 x 25

d = 480 – 475

d = 5

The results present that d = H. I am going to use formula #2 to any extent further.

I shall go back to be able to using a 10 times 10 grid. I should test the formula making use of different sized boxes. We shall draw up the table to record our results. I shall evaluate the outcomes to see in case there is a design forming. This would prove correlation.

NB: To save time We shall only indicate the particular corner numbers in each box. The other numbers inside the box are irrelevant for these calculations.

3 x 3 box #1

1

3

21

twenty-three

deb = 3 x twenty one – 1 x twenty three

d = 63 – 23

d = forty five

3 times 3 box #2

12

14

thirty two

34

d = fourteen x 32 – 13 x 34

d = 448 – 408

d = 40

3 x 3 package #3

80

80

98

100

d = 80 x 98 – 78 x one hundred

d sama dengan 7840 – 7800

d = forty

Algebraic proof:

+2

+2G

a new

a + 2

a + 2G

a + 2G +2

chemical = (a + 2) (a + 2G) – a (a + 2G + 2)

= a2 + 2aG + 2a + 4-G – a2 – 2aG – 2a

= 4G, where Gary the gadget guy = 10, d sama dengan forty

Thus far my theory is correct. I shall keep on to test it upon boxes of increasing found in size. I feel it is just necessary to show one numeric and one algebraic calculation for people boxes.

4 x some box #1

1

four

31

34

d = 4 back button 31 – 1 times 34

d = 124 – 34

d = 90

Algebraic proof:

+3

+3G

the

a + 3

a + 3G

a & 3G +3

d = (a + 3) (a + 3G) – a (an and up. 3G + 3)

= a2 + 3aG + 3a & 9G – a2 – 3aG – 3a

= 9G, wherever G = 10, m = ninety days

5 x 5 package #1

one

5

forty one

45

d = 5 x 41 – 1 x 45

d sama dengan 205 – 45

d = 160

Algebraic resistant:

+4

+4G

a

the + + 4

a + 4G

an and up. 4G +4

d = (a & 4) (a + 4G) – a (a & 4G + 4)

= a2 and up. 4aG + 4a & 16G – a2 – 4aG – 4a

= 16G, where G = 10, m = 160

The results to date are:

10, 40, 90, 160

At this time there is a pattern, which is easier to recognise in case written:

x

1 times 1

2 x 2

3 x a few

4 x some

5 x 5

m

12

40

90

160

1G

4G

9G

16G

12G

22G

32G

42G

The design shows consecutive square numbers. They are also the particular square of one less than the square sizing (x) that produced them.

So:

d = H (x – 1) two

I have extended the results table and added my predicted results, in red, for the remaining box sizes.

x

1 x just one

2 x a couple of

3 x 3

4 x 4

5 x 5

6 x 6

7 times 7

eight x 8

9 x 9

10 x ten

deb

10

40

90

one hundred sixty

250

360

490

640

810

1G

4G

9G

16G

25G

36G

49G

64G

81G

12G

22G

32G

42G

52G

62G

72G

82G

92G

I shall now see if our predicted results are proper by testing three associated with these box sizes.

6 x six box #1

1

six

51

56

d = G (x – 1) 2

= 10 (6 – 1) 2

= 10 (5) 2

sama dengan 10 (25)

= two hundred and fifty

d = 6 by 51 – 1 by 56

d = 306 – 56

d = 250

8 x 8 box #1

1

7

71

78

d sama dengan G (x – 1) 2

= 10 (8 – 1) 2

sama dengan 10 (7) 2

= 10 (49)

= 490

d = 8 x 71 – 1 x 78

d = 568 – 80

d = 490

10 x ten box #1

1

ten

91

100

d = G (x – 1) 2

= 10 (10 – 1) 2

= 10 (9) 2

= 10 (81)

= 810

d = 10 times 91 – 1 by 100

deb = 910 – one hundred

d sama dengan 810

Typically the tests proved my concept correct. I conclude that this investigation has proved that there is connection between square boxes upon a grid. The master formula is:

d = G (x – 1) 2

To look at further We need to change one of the variables.

5. Grid size, from 12 x 10, to electronic. g. 5 x a few, 7 x 7, nine x 9.

* Shape, from a new square, to e. h. a rectangle, a combination, a T-shape.

I possess chosen to check out rectangle boxes on a 10 x 10 grid. I think that right now there will be a master formula regarding any size rectangle within any size grid. I think that it may be more complex because I will have to take into account the reality that a rectangle offers sides of differing plans, opposed to a sq . having sides all the particular same length.

Rectangles can be used two ways:

Rectangle 2

Rectangle 1

I will call rectangle just one a 4 x 2 ( 4 columns by 2 rows) and rectangular shape 2 a 2 times 4 ( 2 columns by 4 rows).

There are several different possible sizes of rectangle within a ten x 10 grid:

2 x 10

2 x nine

a couple of x 8

2 x 7

2 x 6

2 by five

two x 4

2 x 3

3 x 12

3 x nine

3 x 8

3 x 7

3 x six

3 times five

3 x 4

4 x 10

4 x nine

4 by eight

some x 7

4 x 6

4 x a few

5 times ten

5 x 9

5 x 8

5 x seven

5 times six

six x 10

6 x 9

6 x eight

6 times 7

several x 10

7 x 9

7 x 7

8 x ten

7 x 9

9 x 10

This list exhibits 36 possible sizes. If the orientation of the particular rectangle produces a different value for d, and then there will be seventy two possible sizes.

My first task is to ascertain if the orientation of the rectangle impacts the diagonal difference.